%I #23 Jun 25 2021 03:29:28
%S 1,2,3,4,5,6,7,9,10,11,13,14,15,22,23,26,27,28,29,30,39,45,46,47,51,
%T 58,59,61,62,63,86,87,93,94,95,118,119,122,123,124,125,126,147,148,
%U 158,159,178,179,187,188,189,190,214,215,221,222,236,237,238,245,246,247,248,249,253,254
%N Numbers m such that the maximal value of A001222(binomial(m,k)) and the central value A001222(A001405(m)) are identical.
%C Indexes of 0's in A048622. - _Sean A. Irvine_, Jun 24 2021
%H Ivan Neretin, <a href="/A048627/b048627.txt">Table of n, a(n) for n = 1..1000</a>
%e For m=23, A001222 for binomial(23,k) = {0,1,2,3,4,4,5,5,6,6,6,6,6,6,6,6,5,5,4,4,3,2,1,0}, thus both the maximal and central values are 6, so 23 is a term.
%t Select[Range[120], Function[n, ar = PrimeOmega[#] & /@ Binomial[n, Range[0, n/2]]; Max[ar] == ar[[-1]]]] (* _Ivan Neretin_, Sep 07 2015 *)
%o (PARI) isok(m) = vecmax(apply(bigomega, vector(m+1, k, binomial(m,k-1)))) == bigomega(binomial(m, m\2)); \\ _Michel Marcus_, Jun 25 2021
%Y Cf. A001222, A020731, A048622.
%K nonn
%O 1,2
%A _Labos Elemer_